Using I = I0 e^(-mu x), what thickness x is needed to reduce intensity by a factor of 100 if mu = 0.5 cm^-1?

Using I = I0 e^(-mu x), what thickness x is needed to reduce intensity by a factor of 100 if mu = 0.5 cm^-1?

• A. 9.21 cm
• B. 7.00 cm
• C. 18.4 cm
• D. 4.60 cm

Answer: (A)

Attenuation is governed by the exponential decrease I = I0 e^{-μx}. To cut the intensity by a factor of 100, set I/I0 = 1/100 = e^{-μx}. Taking natural logs gives -μx = ln(1/100) = -ln(100), so μx = ln(100). Therefore x = ln(100)/μ. With μ = 0.5 cm^-1, x = 4.60517 / 0.5 ≈ 9.21 cm. For context, at 7 cm the factor is e^{-3.5} ≈ 0.030, about 1/33; at 4.60 cm it's e^{-2.3} ≈ 0.10, about 1/10; at 18.4 cm it's e^{-9.2} ≈ 1e-4, about a factor of 10,000. The required thickness is about 9.21 cm.

Attenuation is governed by the exponential decrease I = I0 e^{-μx}. To cut the intensity by a factor of 100, set I/I0 = 1/100 = e^{-μx}. Taking natural logs gives -μx = ln(1/100) = -ln(100), so μx = ln(100). Therefore x = ln(100)/μ. With μ = 0.5 cm^-1, x = 4.60517 / 0.5 ≈ 9.21 cm.

For context, at 7 cm the factor is e^{-3.5} ≈ 0.030, about 1/33; at 4.60 cm it's e^{-2.3} ≈ 0.10, about 1/10; at 18.4 cm it's e^{-9.2} ≈ 1e-4, about a factor of 10,000. The required thickness is about 9.21 cm.